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3 years ago

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Find the variance and standard deviation of 5,6,7,8,9

1 answer:

34kurt3 years ago

5 0

Step-by-step explanation:

To figure out the variance, subtract 82.5 from the mean, which is 5.5 and then divide by N, which is the value of numbers, (in this case 10) minus 1. The result is a variance of about 9.17. Standard deviation is the square root of the variance so that the standard deviation would be about 3.03.Calculate the mean.

Subtract the mean from each observation.

Square each of the resulting observations.

Add these squared results together.

Divide this total by the number of observations (variance, S2).

Use the positive square root (standard deviation, S).

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No solution because 3=-1 isn’t true.

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Find the values of x for which the combined function h(x) = is undefined if f(x) = 5x + 1 and g(x) = x2 - 9.

Rama09 [41]

We need to know the definition of the "combined function" h(x). I'm going to guess--by looking at answers--that the function is $h(x)=\frac{f(x)}{g(x)}$ making it

$h(x)=\frac{5x+1}{x^2-9}$

This function is undefined wherever the denominator is equal to 0 (division by zero is undefined). Factor the denominator.

$h(x)=\frac{5x+1}{(x+3)(x-3)}$

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3 years ago

Trapezoid JKLM is shown below.

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B is the answer because you have to math it

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3 years ago

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

<u>apply chain rule</u>

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$ <u />

<u>maximum volume:</u> <u>when h = 40/3</u>

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

<u>minimum volume:</u> <u>when h = 0</u>

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

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